Answer :
Let [tex]X[/tex] denote the number of times Joe misses the chute. The distribution for [tex]X[/tex] is binomial with [tex]n=500[/tex] trials and success probability [tex]p=0.2[/tex] (successful in regards to missing the chute).
You then know that the average number of times that he misses is [tex]np=500\times0.2=100[/tex], with a standard deviation of [tex]\sqrt{np(1-p)}\approx8.944[/tex]. These parameters will come in handy in just a moment.
You're asked to compute [tex]\mathbb P(X<80)[/tex]. With the continuity correction, this is approximated by
[tex]\mathbb P(X<80)\approx\mathbb P(X<79.5)=\mathbb P\left(\dfrac{X-100}{8.944}<\dfrac{79.5-100}{8.944}\right)=\mathbb P(Z<-2.292)\approx0.011=1.1\%[/tex]
You then know that the average number of times that he misses is [tex]np=500\times0.2=100[/tex], with a standard deviation of [tex]\sqrt{np(1-p)}\approx8.944[/tex]. These parameters will come in handy in just a moment.
You're asked to compute [tex]\mathbb P(X<80)[/tex]. With the continuity correction, this is approximated by
[tex]\mathbb P(X<80)\approx\mathbb P(X<79.5)=\mathbb P\left(\dfrac{X-100}{8.944}<\dfrac{79.5-100}{8.944}\right)=\mathbb P(Z<-2.292)\approx0.011=1.1\%[/tex]